Respuesta :
The neutralization reaction between potassium hydroxide and sulfuric acid is as follows
2KOH + H2SO4 ---> K2SO4 + 2H2O
number of moles of KOH= (43.74 x 0.500)/ 1000= 0.02187 moles
the reacting ratio of KOH to H2SO4 is 2:1 therefore the moles of H2SO4 is = 0.021187/2= 0.01094 moles
concentration(molarity) = ( 0.01094/50 ) x 1000= 0.2188M
2KOH + H2SO4 ---> K2SO4 + 2H2O
number of moles of KOH= (43.74 x 0.500)/ 1000= 0.02187 moles
the reacting ratio of KOH to H2SO4 is 2:1 therefore the moles of H2SO4 is = 0.021187/2= 0.01094 moles
concentration(molarity) = ( 0.01094/50 ) x 1000= 0.2188M
Answer: The concentration of sulfuric acid is 0.219 M
Explanation:
To calculate the concentration of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is KOH.
We are given:
[tex]n_1=2\\M_1=?M\\V_1=50.00mL\\n_2=1\\M_2=0.500M\\V_2=43.74mL[/tex]
Putting values in above equation, we get:
[tex]2\times M_1\times 50.00=1\times 0.500\times 43.74\\\\M_1=\frac{1\times 0.500\times 43.74}{2\times 50.00}=0.219M[/tex]
Hence, the concentration of sulfuric acid is 0.219 M